Normalised absolute values on $p$-adic extensions

I have the following problem: show that if $L/K$ are finite extensions of the p-adic numbers $\mathbb{Q}_p$ with normalised absolute values $|\cdot|_K$ and $|\cdot|_L$, with $n=[L:K]$, then $|x|_L=|x|_K^n$ $\forall$ $x \in K$.

The intention is to deduce from this that $|N_{L/K}(x)|_K = |x|_L$, but it's the first part I'm having trouble with. I think the issue is I don't know what 'normalised' means in this context: I know a normalised valuation has image $\mathbb{Z}$, but what does it mean in this absolute value situation? Something to do with $|p|_K$ and $|p|_L$ perhaps? I expect the problem is probably quite easy after that: something to do with just defining a basis of size $n$ in K for L and then playing around with the absolute value definition. Thank you for the help!

-
The usual definition of "normalized" is that a uniformizer has absolute value $1/q$, where $q = \# \mathcal{O}_K/\mathfrak{p}_K$ is the cardinality of the residue field. Also, the question as posed does not agree with the conventional absolute values on $p$-adic extensions: typically one sets things up so that $| x |_L = | x |_K$ for $x \in K$, i.e. $| \cdot |_L$ actually extends $| \cdot |_K$. – Justin Campbell Nov 13 '11 at 23:14
Does that definition of "normalized" make this problem correctly formulated then? I was thinking that myself - obviously since K contains $\mathbb{Q}_p$ if the absolute values don't agree on K then they can't both extend the conventional absolute value. I was hoping after knowing how the question was framed it would become somewhat obvious, but unfortunately not! – Ben Nov 13 '11 at 23:29
Yes, it's correctly formulated, so we see that the normalized absolute value is actually a power of the extended absolute value. The key point is that $n = ef$, where $e$ is the ramification degree and $f$ is the degree of the residue field extension. If this is not enough of a hint I can write up the details in an answer. – Justin Campbell Nov 13 '11 at 23:37
The ramification degree of what, sorry? I don't think I'm totally clear on the solution; if it's not too much trouble, would you mind writing up the details? – Ben Nov 14 '11 at 1:52
@Ben There are only two primes in sight, so if we call the uniformizers $\pi_K$ and $\pi_L$ then $e$ should be such that $\pi_K = u\pi_L^e$ for some unit $u$ of $\mathscr{O}_L$. – Dylan Moreland Nov 14 '11 at 2:14

There are various ways to normalize the absolute value on a local field. As Justin Cambell mentions, one way is to choose the absolute value so that it agrees with a given absolute value on some natural choice of subfield (e.g. with the usual absolute value on $\mathbb R$, in the case of $\mathbb C$ over $\mathbb R$, or with the $p$-adic absolute value on $\mathbb Q_p$, in the case of an extension of $\mathbb Q_p$).

However, there is a more intrinsic way to normalize the absolute value, which does not make any reference to a given choice on a subfield. Namely, if $x$ is any non-zero element of the local field $K$, then multiplication by $x$ scales additive Haar measure on $K$ by some amount; we declare this scaling factor to be $|x|$. This is the normalized absolute value on $K$ referred to in your question.

E.g. if $K = \mathbb R$, this will give the usual absolute value on $\mathbb R$. If $K = \mathbb C$, it will give the square of the usual absolute value on $\mathbb C$.

If $K$ is an extension of $\mathbb Q_p$ with residue field of order $q$ and $\pi$ is a uniformizer, then (by comparing the measure of $\mathcal O_K$ and $\pi \mathcal O_K$) one finds that $|\pi| = q$.

Now with this normalization, your problem is fairly easily solved.

One approach is to use the relation $[L:K] = e f$, as Justin Cambell suggests.

An alternative approach is to use directly the fact that $L$ is a vector space over $K$ of dimension $[L:K]$, and think about how Haar measure on $L$ is thus related to Haar measure on $K$, and how it will scale under multiplication by an element of $K$. (The analogy with the case of $\mathbb C$ over $\mathbb K$ can be helpful here.)

Note that using the second approach, and then considering the first approach in reverse, one can derive the formula $[L:K] = ef$. This is a typical application of normalized absolute values. (I.e. one begins with a more analytic viewpoint, and then uses it to derive arithmetic facts.) Another is to derive the product formula for a global field (that the product over all absolute values of an element of the global field equals $1$); and note that in this formula, it is crucial that normalized absolute values be used.

-
For reference: the Haar measure approach that Prof Emerton mentions is adopted from the beginning in Weil's book. – Dylan Moreland Nov 14 '11 at 1:34